Question: $h'(x)=-h(x)$, and $h(0)=12$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $h(x)=-12e^{x}$ (Choice B) B $h(x)=12e^{-x}$ (Choice C) C $h(x)=-e^{x}$ (Choice D) D $h(x)=-12e^{-x}$
The general solution of equations of the form $h'(x)=kh(x)$ is $h(x)=C\cdot e^{kx}$ for some constant $C$. This can be found using separation of variables. In our case, $k=-1$, so $h(x)=C\cdot e^{-x}$. Let's use the fact that $h(0)=12$ to find $C$ : $\begin{aligned} h(x)&=C\cdot e^{-x} \\\\ h(0)&=C\cdot e^{-1\cdot 0} \gray{\text{Plug }x=0} \\\\ 12&=C\cdot e^{-1\cdot 0} \gray{h(0)=12} \\\\ 12&=C \end{aligned}$ In conclusion, $h(x)=12e^{-x}$.